R E S E A R C H Open Access
Stability regions of discrete linear periodic systems with delayed feedback controls
Jong Son Shin1, Rinko Miyazaki1,2and Dohan Kim3*
*Correspondence:[email protected]
3Department of Mathematics, Seoul National University, Gwanak-gu, Seoul 08826, Korea
Full list of author information is available at the end of the article
Abstract
We propose a geometric method to determine the stability region of the zero solution of a linear periodic difference equation via the delayed feedback control (briefly, DFC) with the commuting feedback gain. For the equation, our method is more effective than the Jury criterion. First, we give a relationship, named theC-map theorem, between the characteristic multipliers of an original equation and those of the equation via DFC. Next, we show the existence andm-starlike property, defined in this paper, of anm-closed curveinduced from the C-map. Using this result, we prove that the region enclosed by them-closed curve is the stability region of the zero solution of the equation via DFC.
Mathematics Subject Classification: Primary 39A06; 39A30; secondary 93B52 Keywords: Stability region; Periodic linear difference equation; Delayed feedback;
C-map theorem; m-closed curve; m-starlike curve
1 Introduction and preliminaries 1.1 Introduction
The delayed feedback control (DFC) is an important method for stabilizing the unstable periodic orbitφ(t) with periodω> 0 to a differential equation
x(t) =f(x), x∈⊂Rd, (E)
embedded within a chaotic attractor. As DFCs, Pyragas [11] has firstly used a perturbation u(t) =K(x(t–ω) –x(t)) to Equation (E), that is,
x(t) =f x(t)
+K
x(t–ω) –x(t)
, (DF)
where ad×dreal constant matrixKis the so-called feedback gain, and he numerically determined the feedback gainK so that the periodic solution of Equation (DF) is sta- ble.
To stabilize theoretically the unstable periodic orbit, Miyazaki, Naito, and Shin [8] have used the method of linearization under the commuting condition for the gainK. Then,
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the linear variational equations around the orbitφ(t) for Equation (DF) becomes y(t) =A(t)y(t) +K
y(t–ω) –y(t)
, (LDF)
whereA(t) =Df(φ(t)) is the Jacobian off(x).
A discrete version of Equation (DF) is given by the form x(n+ 1) =f
x(n)
+u(n), u(n) =K
x(n–ω) –x(n)
,n∈Z:={0,±1,±2, . . .}, whereω∈Z∞1 :={1, 2, . . .}.
This type of feedback scheme has certain inherent limitations [12]. On the other hand, Buchner and Żebrowski [1] considered a perturbation of the echo-type formulated as u(n) =K(x(n–ω+ 1) –A(n)x(n)) to study the stability and the bifurcation for the logistic map. This method is considered as a prediction-based feedback control [13] or nonlinear feedback control [14]. For other types ofu(n) see [15–17]. Furthermore, Ohta, Takahashi, and Miyazaki [9] made a remark that DFC of the echo type is more effective than Pyragas type for one-dimensional case.
As the first step of the study, we are interested in the problem of stabilizing the unstable zero solution to linear periodic difference equations of the form
x(n+ 1) =A(n)x(n), n∈Z, (L)
apart from nonlinear difference equationsx(n+ 1) =f(x(n)).
Here we assume thatA(n) is ad×dcomplex matrix with periodωandx(n) belongs to theddimensional complex Euclidean spaceCd.
In this paper, we adopt the perturbation of the echo type and consider the following equation with DFC
y(n+ 1) =A(n)y(n) +K
y(n–ω+ 1) –A(n)y(n)
. (LF)
The goal of the paper is to describe the stability region, containing all the characteristic multipliers of Equation (L), of the zero solution to Equation (LF) for general periodω≥3 (refer to [7] forω= 2). We develop ageometric methodto characterize the stability region.
As a next step, for periodic solutions with periodω, we will investigate the stability region in the forthcoming paper [5], whose main results rely strongly on this paper.
In general, the stability of the zero (or periodic) solution of Equation (LF) is determined by the absolute values of these characteristic multipliers, i.e., the roots of its characteristic polynomial. However, for the characteristic polynomial of Equation (LF), it is very difficult to apply the classical criteria of Schur–Cohn or Jury in [4] as well as to determine the stability region, since they are based on algebraic methods. Indeed, the order of the inner matrix becomes very large as the dimensiondand the periodωincrease. For example, according to our experimental calculation, the criteria of Schur–Cohn or Jury are very complicated even for the case whenω= 4 andd= 1. This is a motivation for this paper.
To solve such a difficulty, we introduce a new geometric method. As a main result, we can theoretically determine the stability region in general whenK=kE. In particular, when ω= 4 and all the characteristic multipliers of Equation (L) are real, our method can give a more concrete and precise stability region (Fig.2).
Our geometric method is developed as follows.
First, we establish a relationship, named the C-map theorem (Theorem2.5and Corol- lary2.6), between characteristic multipliers of Equation (L) and Equation (LF). To carry out this, we introduce a C-map under the commuting condition:KA(n) =A(n)Kfor all n∈ {0, 1, 2, . . . ,ω– 1}, which is motivated by the paper [8] for a continuous system. For ex- ample, for a characteristic multiplierμof Equation (L) and for a characteristic multiplier νof Equation (LF) withK=kE,ka real number andEthe identity matrix, the C-map is given byμ=Cω,k(ν) =ν((1–k)νν–k )ω.
Next, we give geometric properties of the imageBω,k(θ) :=Cω,k(eiθ),θ∈(–π,π] by the C- map of the unit circle in the complex plane. In general, the above image is not geometrically simple. We show the existence and them-starlike property (Theorem6.4) of anm-closed curve (Definition6.3) as a part of the imageBω,k(θ).
Finally, using this result, we prove that if all the characteristic multipliers of Equation (L) are in the interior of the (stability) region enclosed by anm-closed curve, then the zero solution to Equation (LF) is asymptotically stable (Theorem7.2). Furthermore, we give necessary and sufficient conditions for all the characteristic multipliers of Equation (L) to be in the interior of the region. Our method is illustrated for the cases whenω= 3, 4 and all the characteristic multipliers of Equation (L) are real.
The paper is organized as follows.
Section1. Introduction and preliminaries
Section2. Characteristic multipliers for Equation (LF) Section3. Properties of the functionBω,k(θ)
Section4. Existence of solutions of EquationBω,k(θ) = 0 Section5. EquationBω,k(θ) = 0
Section6. Geometric properties of the functionBω,k(θ) Section7. Stability regions
1.2 Preliminaries
In this subsection, we give some basic properties of the characteristic multipliers for Equa- tion (L) and Equation (LF). LetXbe a Banach space withdimX<∞andL:X→Xa bounded linear operator. We denote byN(L) the null space ofL, and byWη(L) andGη(L) the eigenspace and the generalized eigenspace forη∈σ(L), respectively, whereσ(L) stands for the set of all eigenvalues ofL. LetZ∞p ={p,p+ 1, . . .}forp∈Z. For anym,n∈Zwith m<nwe setZnm={m,m+ 1, . . . ,n– 1,n}.
First, we consider Equation (L), which has the matrix coefficientA(n) with periodω.
Throughout this paper we assume that (A):A(n) is nonsingular for alln∈Zω–10 .
Then the unique solutionx(n;m,x0) of Equation (L) through the initial point (m,x0)∈ Z×Cdis given byx(n;m,x0) =T(n,m)x0, whereT(n,m),n,m∈Zstands for the solution operator of Equation (L). SetT(n) =T(n+ω,n),n∈Z. ThenT(0) is called the periodic operator of Equation (L). ThenT(n,m) (m,n∈Z) andT(0) are given by
T(n,m) =
n–1
i=m
A(i)(n≥m) and T(0) =
ω–1
i=0
A(i),
respectively, where n–1
i=m
A(i) =
⎧⎨
⎩
A(n– 1)A(n– 2)· · ·A(m) (n>m),
E (n=m).
ThusT(n,m),n,m∈Zhas following properties (refer to [2,10]):
(T1)T(n,n) =E,n∈Z.
(T2)T(n,m)T(m,r) =T(n,r),m∈Znr. (T3)T(n+ω,m+ω) =T(n,m),m≤n.
Note that usingω–periodicity ofA(n),
T(1) =A(0)T(0)A(0)–1. (1)
A complete study of (L) is carried out by the so-calledFloquet theory(see, for example, C. Pötzsche [10]) Note thatσ(T(n)) =σ(T(0)) andT(0) is nonsingular by Condition (A).
Thus 0 /∈σ(T(0)). From now on,μ∈σ(T(0)) is called theFloquet’s multiplierorcharacter- istic multiplierof Equation (L) (refer to [2,10]). We recall that the location of eigenvalues ofT(0) determines the stability properties of Equation (L).
Next, we consider Equation (LF). LetCω–1 be the set of all maps fromZ0–ω+1 intoCd, which is the Banach space equipped with the norm|ϕ|Cω–1=sups∈Z0
–ω+1|ϕ(s)|. It is obvious thatdimCω–1=ωd. Letm∈Zbe fixed. For any functiony:Z∞m–ω+1→Cdand anyn∈ Z∞m, we define a functionyn:Z0–ω+1→Cdbyyn(s) =y(n+s),s∈Z0–ω+1. For anyn∈Z∞m the unique solutionyn(m,ϕ)∈Cω–1 of Equation (LF) through the initial point (m,ϕ)∈ Z×Cω–1 is given byyn(m,ϕ) =UK(n,m)ϕ, whereUK(n,m) :Cω–1→Cω–1 stands for the solution operator of Equation (LF). SetUK(n) =UK(n+ω,n),n∈Z. ThenUK(0) is called the periodic operator of Equation (LF). Hereafter, ifK=kE, then we denote byUk(n,m) andUk(0) the operatorsUK(n,m) andUK(0), respectively.
The following result can be proved by a similar argument as in the proof of [3, p. 237, Lemma 1.1].
Proposition 1.1 νis a characteristic multiplier of Equation(LF)if and only if there is a nontrivial solution yn,n∈Z∞0 of Equation(LF)of the form
y(n+ω) =νy(n), n∈Z∞–ω+1. (2)
Hereafter, we assume the following condition (K) for the feedback gainK:
(K-1)σ(K)⊂R,
(K-2) 0 <|κ|< 1 for allκ∈σ(K), (K-3)σ(UK(0))∩σ(K) =∅.
Ifk∈Rwith 0 <|k|< 1, thenK=kEsatisfies the condition (K) (see Lemma1.3for a proof ). However, the condition (K-3) does not hold for a general matrixK, while it can be replaced by other conditions (see [6]).
Now, we introduce the commuting condition (C).
(C)KA(n) =A(n)K, (n∈Z).
The proof of the following lemma is easy.
Lemma 1.2 For Equation(L)the following statements are equivalent:
(1)A(n)K=KA(n),n∈Z.
(2)T(n,m)K=KT(n,m),n,m∈Z. (3)T(n, 0)K=KT(n, 0),n∈Z.
For the caseK=kE, the following result holds.
Lemma 1.3 Let K=kE, 0 <|k|< 1,and k∈R.If Condition(A)is satisfied,then Conditions (C)and(K)are satisfied.
Proof SinceK=kE, we obtain that the condition (C) is clearly satisfied. Moreover,σ(K) = {k}andWk(K) =Cd. Now, we show by contradiction that the condition (K-3) is satisfied.
Supposek∈σ(Uk(0)). Then there exists a nontrivial solutiony(n) of Equation (LF) such thaty(n+ω) =ky(n),n∈Z–ω+1by Proposition1.1. Hencey(n+ 1 –ω) =k–1y(n+ 1),n∈Z. Substituting this relation into Equation (LF), we havey(n+ 1) =A(n)y(n) +k[k–1y(n+ 1) – A(n)y(n)], which implies that (1 –k)A(n)y(n) = 0. Sincek= 1 andA(n) is nonsingular, we havey(n) = 0. This leads to a contradiction, sincey(n) is a nontrivial solution. Hence, the
condition (K-3) is satisfied.
Hereafter, we always assume Conditions (A), (K), and (C) in this paper.
We note that under the condition (2), Equation (LF) becomes y(n+ 1) =K(ν)–1A(n)y(n),
where
K(ν) =ν–1(νE–K)(E–K)–1. (3)
Finally, we will transform Equation (LF) to the extended linear periodic difference equa- tion. By transforming
y
n– (ω– 1)
=z(1;n), y
n– (ω– 2)
=z(2;n), . . . , y(n) =z(ω;n)
in Equation (LF) and settingz(n) =t(tz(1;n),tz(2;n), . . . ,tz(ω;n)), Equation (LF) becomes
z(n+ 1) =BK(n)z(n), (BE)
where
BK(n) =
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎝
0 E 0 · · · 0 0
0 0 E · · · 0 0
0 0 0 · · · 0 0
... ... ... ... ... ... ... ... ... ... ... 0 ... ... ... ... . .. 0 0 · · · 0 E K 0 0 · · · 0 (E–K)A(n)
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎠
, (4)
which is called theextended feedback equationof Equation (LF).
Then the following result is easy to prove.
Lemma 1.4 detBK(n) = (–1)(ω–1)ddetK for all n∈Z∞0 . Proof An easy calculation yields
detBK(n) = (–1)(ω–1)ddet
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎝
E 0 · · · 0 0 0
0 E · · · 0 0 0
... ... . .. ... ... ...
0 0 · · · E 0 0
0 0 · · · 0 E 0
0 0 · · · 0 (E–K)A(n) K
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎠
= (–1)(ω–1)ddet
E 0 (E–K)A(n) K
= (–1)(ω–1)ddetK.
This completes the proof.
It follows from Lemma1.4that if 0 /∈σ(K), then the existence and uniqueness of solu- tions to Equation (BE) is guaranteed. We denote byTB(n,m) andTB(0) the solution oper- ator and the periodic operator of Equation (BE), respectively. LetC:=C1andRstand for the set of all the real numbers.
Now, we give a relationship between the operatorsUK(0) andTB(0).
Define a mappingSω–1fromCω–1intoCωd:=
ω
Cd×Cd× · · · ×Cdby
ϕ∈Cω–1→tt
ϕ(–ω+ 1), tϕ(–ω+ 2), . . . ,tϕ(–1), tϕ(0)
∈Cωd.
ThenSω–1is bijective. Hence, we haveSω–1UK(n,m)ϕ=TB(n,m)Sω–1ϕ.
Indeed, we have
Sω–1UK(n,m)ϕ=Sω–1yn(m,ϕ)
=
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎝
y(n– (ω– 1);m,ϕ) y(n– (ω– 2);m,ϕ)
... y(n– 1;m,ϕ)
y(n;m,ϕ)
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎠
=
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎝ z(1;n) z(2;n)
... z(ω– 1;n)
z(ω;n)
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎠
=TB(n,m)
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎝
ϕ(–ω+ 1) ϕ(–ω+ 2)
... ϕ(–1)
ϕ(0)
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎠
=TB(n,m)Sω–1ϕ.
SoUK(n,m) is uniquely extended ton<mas follows:
UK(n,m) =S–1ω–1TB(n,m)Sω–1, n<m.
From this we have
Sω–1UK(n,m) =TB(n,m)Sω–1 (m,n∈Z), Sω–1UK(0) =TB(0)Sω–1. SinceUK(0) andTB(0) are similar, the following relations hold:
UK(0)ϕ=νϕ ⇐⇒ Sω–1UK(0)ϕ=νSω–1ϕ ⇐⇒ TB(0)Sω–1ϕ=νSω–1ϕ.
Therefore, we obtain the following result.
Lemma 1.5 σ(UK(0)) =σ(TB(0))and0 /∈σ(UK(0)).
Proof Combining Lemma1.4and the condition (K-2), we havedetBK(n)= 0. SinceUK(0) andTB(0) are similar,σ(UK(0)) =σ(TB(0)) and hence 0 /∈σ(UK(0)).
2 Characteristic multipliers for Equation (LF)
In this section, we determine the spectrumσ(UK(0)) of the periodic operator of Equation (LF) and establish theC-map theorem.
2.1 Spectrum of the periodic operatorUK(0) Set
Hmn= (E–K)n–mT(n,m), n≥m.
ThenHmn has the following properties:
Hkk=E, HknHmk =Hmn, (E–K)A(n)Hmn=Hmn+1. (5) Indeed, using the commuting condition (C) and Lemma1.2, we have
(E–K)A(n)Hmn= (E–K)n+1–mA(n)T(n,m)
= (E–K)n+1–mT(n+ 1,m) =Hmn+1.
Inductively, we can obtain a representation ofTB(0) as follows:
TB(0) =
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎝
K 0 0 · · · 0 H01
KH12 K 0 · · · 0 H02
... ... ... ... ... ...
KH1ω–1 KH2ω–1 KH3ω–1 · · · K H0ω–1 KH1ω KH2ω KH3ω · · · KHω–1ω H0ω+K
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎠ .
Now, we will calculatedet(TB(0) –νE).
Proposition 2.1 The characteristic polynomial of TB(0)is given as follows:
det
TB(0) –νE
=det
(–1)ωνω–1(E–K)ω det
νK(ν)ω–T(0) . In particular,det(TB(0) –νE) = 0if and only ifdet(νK(ν)ω–T(0)) = 0.
Proof Set
M=
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎝ E
–H12 E
0
–H23 E . .. ...
. .. . ..
0
–Hω–2ω–1 E –Hω–1ω E⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎠ .
ThendetM= 1. Under the condition (C), by Schur’s formula, we have det
TB(0) –νE
=det M
TB(0) –νE
=det
M11 M12
M21 M22
=detM22det
M11–M12M–122M21 , where
M11=
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎝ K–νE
νH12 K–νE
νH23 K–νE
0
. .. . ..
0
. .. . ..νHω–2ω–1 K–νE
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎠
, M12=
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎝ H01
0 ... ... ... 0 0
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎠ ,
M21=
0 0 · · · 0 νHω–1ω
, M22=K–νE.
Here, we have used the condition (K-3) and the formula for the determinant of a block matrix with four submatrices. Thus we have
det
TB(0) –νE
=det(K–νE)
×det
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎜⎝
K–νE –ν(K–νE)–1H01Hω–1ω
νH12 K–νE 0
νH23 K–νE
0
0. .. . .. ...
0
. .. . ..K–νE 0
νHω–2ω–1 K–νE
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎟⎠
...
=det(K–νE)ω–2det
K–νE (–ν)ω–2{(K–νE)–1}ω–2H01H2ω νH12 K–νE
. (6)
It follows from (6) that det
TB(0) –νE
=det(K–νE)ω–1det
K–νE+ (–ν)ω–1
(K–νE)–1ω–1
H01H1ω
=det
(K–νE)ω+ (–ν)ω–1H01H1ω
=det
(K–νE)ω+ (–ν)ω–1(E–K)ωT(1, 0)T(ω, 1)
=det
(K–νE)ω+ (–ν)ω–1(E–K)ωT(1) by the property (T3) SinceT(1) =A(0)T(0)A–1(0) andA(0)K=KA(0) hold, we have
det
TB(0) –νE
=det
(K–νE)ω+ (–ν)ω–1(E–K)ωT(1)
=det
(K–νE)ωA(0)A(0)–1+ (–ν)ω–1(E–K)ωA(0)T(0)A(0)–1 (by(1))
=det A(0)
(K–νE)ω+ (–ν)ω–1(E–K)ωT(0) A(0)–1
=det
(–1)ωνω–1(E–K)ω ν
ν–1(E–K)–1ω
(νE–K)ω–T(0)
=det
(–1)ωνω–1(E–K)ω det
νK(ν)ω–T(0) .
Since ν = 0 by Lemma 1.5 and 1 /∈σ(K), we obtain that if det(TB(0) –νE) = 0, then
det(νK(ν)ω–T(0)) = 0, and vice versa.
Combining Proposition2.1and Lemma1.5, we obtain the following equivalence.
Proposition 2.2 The following statements are equivalent:
(1)ν∈σ(UK(0)).
(2)det(TB(0) –νE) = 0.
(3)det(νK(ν)ω–T(0)) = 0.
Theorem 2.3 The following statements hold.
(1)Letν∈σ(UK(0)).Thenψ∈Wν(UK(0))⇐⇒Sω–1ψ∈Wν(TB(0)).
(2)The characteristic equationdet(TB(0) –νE) = 0hasωd roots.
Proof We prove only the assertion (1). Assumeψ∈Wν(UK(0)). SinceUK(0)ψ=νψ, we have Sω–1UK(0)ψ =Sω–1νψ. Since Sω–1UK(0)ψ =TB(0)Sω–1ψ, we obtainTB(0)Sω–1ψ = Sω–1νψ, that is,Sω–1ψ∈Wν(TB(0)), and vice versa.
Combining Theorem2.3with Lemma1.4, we obtain the following result.
Proposition 2.4 Letν1, . . . ,νωd,counted with multiplicity,be all the characteristic multi- pliers of Equation(LF).Thenν1· · ·νωd= (detK)ω.
Proof Combining Theorem2.3with Lemma1.4, we obtain ν1· · ·νωd=detTB(0) =
ω–1 n=0
detBK(n)
=
(–1)(ω–1)ddetKω
= (–1)ω(ω–1)d(detK)ω.
Sinceω(ω– 1)dis an even number, the proof is complete.
It follows from Proposition2.4that (1) ifK=kE, then
ν1ν2· · ·νωd= (k)ωd;
(2) ifk1,k2, . . . ,kd, counted with multiplicity, are eigenvalues of the matrixK, thendetK= k1k2· · ·kdand
ν1· · ·νωd= (detK)ω= (k1k2· · ·kd)ω.
This implies that ifdetK> 1, then there exists aνi∈σ(UK(0)) such that|νi|> 1. In other words, the zero solution of Equation (LF) is unstable ifdetK> 1. Note thatdetK> 1 if K=kEand|k|> 1.
2.2 C-map theorems
In this subsection, we introduce the C-map Theorems, which give the relationship be- tween the characteristic multipliers of Equations (L) and (LF) and play the crucial role throughout this paper. For commuting matricesAandBwe set
σ[AB] =
(α,β)∈σ(A)×σ(B)|αβ∈σ(AB) ,
whereσ(AB) ={αβ|α∈σ(A),β∈σ(B),Wα(A)∩Wβ(B)=∅}.
For a functionf(x,y), we denote byfx(y) andfy(x) the functionf(x,y) ofyfor each fixed x, and the functionf(x,y) ofxfor each fixedy, respectively.
In view ofK(ν)ωin Proposition2.1, we introduce g(k,z) =
z–k (1 –k)z
ω
:I×D→C\ {0} and Cω,k(z) =zg(k,z),
whereI=R\ {1}andD=C\R. The functionCω,k(z) is called the characteristic multiplier map (briefly,C-map) for Equation (LF). Note thatg(K,z) is well defined andzg(K,z) is nonsingular for allz∈D, sinceg(k,z) is analytic inkfor allz∈D.
We are now in a position to state and prove the C-map theorem for Equation (LF).
Theorem 2.5 (C-map Theorem) ν ∈ σ(UK(0)) if and only if there exists a (k,μ) ∈ σ[KT(0)]such thatμ=Cω,k(ν).
Proof It follows from Proposition 2.2 that ν ∈σ(UK(0)) if and only if det(νK(ν)ω – T(0)) = 0, that is, 0∈σ(νg(K,ν) –T(0)). According to the spectral mapping theorem, we haveσ(νg(K,ν)) ={νg(k,ν)|k∈σ(K)}. Moreover, it follows from Condition (C) and [8, Lemma 4.1] thatνg(K,ν) andT(0) commute. Therefore, by [8, Lemma A.1], the condi- tion 0∈σ(νg(K,ν) –T(0)) implies thatν∈σ(UK(0)) if and only if there existk0∈σ(K) andμ∈σ(T(0)) such that
μ=νg(k0,ν), Gνg(K,ν)
νg(k0,ν)
∩GT(0)(μ)={0}. (7)
For such ak0∈σ(K), we denote by{k0,k1, . . . ,kp},p≤d– 1 the set ofk∈σ(K) such that νg(k,ν) =νg(k0,ν). Using the spectral mapping theorem again, we haveGνg(K,ν)(νg(k0,ν)) = p
i=0GK(ki). Therefore, we see that Gνg(K,ν)(νg(k0,ν))∩GT(0)(μ) ={0} if and only if GT(0)(μ)∩p
i=0GK(ki)={0}. Then x∈GT(0)(μ)∩p
i=1GK(ki), x= 0 can be expressed asx=p
i=0Pix,Pix∈GK(ki), wherePi:Cd→GK(ki) is the projection. SinceT(0) andK commute, we haveT(0)Pix=PiT(0)x=Piμx=μPix,i= 0, . . . ,p. Since there is at least one isuch thatPix= 0, we have
GK(ki)∩GT(0)(μ)={0}. (8)
It follows from [8, Lemma A.2] that the condition (8) is reduced to the conditionWK(ki)∩ WT(0)(μ)={0}. Hence the condition (7) is replaced by the condition
μ=Cω,ki(ν), WK(ki)∩WT(0)(μ)={0}.
Thus we have (ki,μ)∈σ[KT(0)]. This proves the theorem.
Corollary 2.6 Let K=kE.Thenν∈σ(Uk(0))if and only if Cω,k(ν)∈σ(T(0)).
TheC-mapμ=Cω,k(z) can be reformulated as
Pω,k(z;μ) = (z–k)ω–μ(1 –k)ωzω–1= 0. (9) Using (9) and Corollary2.6, we obtain the following result.
Corollary 2.7 Let K=kE.Then for everyμ∈σ(T(0))the equationμ=Cω,k(ν)hasω solutions,counted with multiplicity,which belong toσ(Uk(0)).
3 Properties of the functionBω,k(θ)
In this section we consider several properties of the image Bω,k(θ) :=Cω,k
eiθ
=
1 –ke–iθ 1 –k
ω
eiθ, –π<θ≤π (10)
by the C-mapCω,k(z) of the unit circle. Clearly, we have:
(1)Bω,k(0) = 1∈R. (2)Bω,k(π) = –(1+k1–k)ω∈R.
(3)Bω,k(θ) is differentiable on [0,π].
Note thatlimk→1|Bω,k(π)|=∞. Hereafter, we assume thatω∈Z∞3
We denote byCthe closed unit disc, i.e.,C={z||z| ≤1}, and denote byn(∂C,Cω,k) the winding number ofCω,k(ν) whenνrotates along the unit circle∂Ccentered at the origin in the positive direction.
Lemma 3.1 The following statements hold.
(1)Bω,k(θ)= 0for allθ∈(–π,π].
(2)Bω,k(θ+ 2nπ) =Bω,k(θ)andBω,k(θ) =Bω,k(–θ), (n∈Z, –π<θ≤π).
(3)Bω,k(–π) :=limθ→–πBω,k(θ) = –(1+k1–k)ω∈R.
(4)Cω,k(1) = 1and Cω,k(ν) =Cω,k(ν).
(5)n(∂C,Cω,k) = 1.
Proof (1), (2), (3), and (4) are obvious. (5) By the argument principle, we have n(∂C,Cω,k) = 1
2πi
∂C
Cω,k(ν)
Cω,k(ν)dν= 1 2πi
∂C
ω ν–kdν+
∂C
1 –ω ν dν = 1
as required.
To obtain a representation ofBω,k(θ), for anyk, 0 <|k|< 1 and anyθ∈(–π,π], we define β(k,θ) as
tanβ(k,θ) = ksinθ
1 –kcosθ, !!β(k,θ)!!<π
2. (11)
Now, we give elementary properties ofβ(k,θ).
Lemma 3.2 Forθ∈(0,π)the following statements hold:
(1) 0 <k< 1if and only if0 <β(k,θ) <π2 for allθ∈(0,π).
(2) –1 <k< 0if and only if–π2 <β(k,θ) < 0for allθ∈(0,π).
(3)For anyθ∈(0,π)β(k,θ)is increasing in k, 0 <|k|< 1.
The assertions (1) and (2) in Lemma3.2imply thatβ(k,θ)= 0 for allθ ∈(0,π). Since
sinβ(k,θ)
cosβ(k,θ)=1–kksinθcosθ, (11) can be replaced by ksin
β(k,θ) +θ
=sinβ(k,θ). (12)
Next, we give two representations ofBω,k(θ). We need the following identity often 1 –ke–iθ=√
1 – 2kcosθ+k2eiβ(k,θ). (13)
Now, the following representation ofBω,k(θ) is given by (13).
Proposition 3.3 Bω,k(θ)can be reformulated as
Bω,k(θ) =(1 – 2kcosθ+k2)ω2
(1 –k)ω eiϕk(θ), θ∈(–π,π], where
ϕk(θ) =ωβ(k,θ) +θ, –π<θ≤π. (14) Corollary 3.4 The following results hold.
(1)β(k, 0) = 0andϕk(0) = 0for all k(0 <|k|< 1).
(2)β(k,π) = 0andϕk(π) =πfor all k(0 <|k|< 1).
(3)β(k,θ)= 0for all k(0 <|k|< 1)andθ(0 <|θ|<π).
Using Proposition3.3, we have
!!Bω,k(θ)!!=(1 – 2kcosθ+k2)ω2
(1 –k)ω (15)
and|ϕk(θ)|< (ω2 + 1)π. Thus the following result holds.
Lemma 3.5 Letθ∈[0,π].Then the following statements hold.
(1)If0 <k< 1,then|Bω,k(θ)| ≥1and|Bω,k(θ)|is strictly increasing inθ. (2)If–1 <k< 0,then|Bω,k(θ)| ≤1and|Bω,k(θ)|is strictly decreasing inθ. Corollary 3.6 The following statements hold.
(1)If0 <k< 1,thenmin0≤θ≤π|Bω,k(θ)|=Bω,k(0) = 1andmax0≤θ≤π|Bω,k(θ)|=|Bω,k(π)|= (1+k1–k)ω.
(2)If–1 <k< 0,thenmin0≤θ≤π|Bω,k(θ)|=|Bω,k(π)|= (1+k1–k)ωandmax0≤θ≤π|Bω,k(θ)|= Bω,k(0) = 1.
Since
!!Bω,k(θ)!!ω2 = 1 + 2k
(1 –k)2(1 –cosθ), (16)
we have the following lemma.
Lemma 3.7 Let0 <|k|< 1andθ∈(0,π].Then|Bω,θ(k)|is strictly increasing in k.
Proof Setb(k) :=|Bω,θ(k)|ω2. In view of (16), we haveb(k) = 2(1+k)(1–k)3(1 –cosθ) > 0, and hence
b(k) is strictly increasing ink.
4 Existence of solutions of equationBω,k(θ) = 0
In this section, we give the criteria for the existence of solutions of EquationBω,k(θ) = 0, i.e.,Bω,k(θ)∈Ron [0,π]. In other words,sinϕk(θ) = 0. Sinceβ(k, 0) =β(k,π) = 0 for any k, 0 <|k|< 1 by Corollary3.4,θ= 0,πare the solutions of EquationBω,k(θ) = 0. Thus we consider the caseθ∈(0,π). To discuss this problem, we investigate separately two cases 0 <|k| ≤ω–11 andω–11 <|k|< 1.
First, we need the following properties of ϕk(θ). Since ϕk(θ) =ωβ(k,θ) +θ, we have
d
dθϕk(θ) = 1–2kζ(k,θ)cosθ+k2, whereζ(k,θ) =k(ω– 2)cosθ– (ω– 1)k2+ 1.
Proposition 4.1 The following statements hold forθ∈[0,π].
(1)ϕk(θ) = 0⇐⇒ζk(θ) = 0, i.e.,cosθ=(ω–1)kk(ω–2)2–1. (2)ϕk(θ) > 0⇐⇒ζk(θ) > 0 ;ϕk(θ) < 0⇐⇒ζk(θ) < 0.
(3)ϕk(θ)is continuous on[0,π].
(4) ω–11 <|k|< 1⇐⇒ |(ω–1)kk(ω–2)2–1|< 1.
Corollary 4.2 The following statements hold.
(1)k= –ω–11 if and only ifϕk(0) = 0.
(2)k=ω–11 if and only ifϕk(π) = 0.
The following result is easily derived from the above argument and Proposition4.1.
Corollary 4.3 Forθ= 0,πthe following statements hold.
(1)The caseθ= 0.
(1-1)If–ω–11 <k< 1,thenϕk(0) > 0.
(1-2)If–1 <k< –ω–11 ,thenϕk(0) < 0.
(2)The caseθ=π.
(2-1)If–1 <k<ω–11 ,thenϕk(π) > 0.
(2-2)If ω–11 <k< 1,thenϕk(π) < 0.
Next, we show that solutions of EquationBω,k(θ) = 0 on [0,π] for the case 0 <|k| ≤ω–11 areθ= 0 andπonly.
We are now in a position to state and prove the first main theorem in this section.
Theorem 4.4 Let θ ∈[0,π]. Suppose 0 < |k| ≤ ω–11 . Then Bω,k(θ) = 0 if and only if θ= 0,π.
Proof The proof is based on Proposition4.1and Corollary4.3.
(1) If 0 <k≤ω–11 , thenϕk(π)≥0 by Corollary4.2and Corollary4.3. Moreover, Propo- sition4.1implies the inequalitycosπ= –1≥ (ω–1)kk(ω–2)2–1. On the other hand, we have the inequalitycosθ>cosπ≥(ω–1)kk(ω–2)2–1 on [0,π). Thusϕk(θ) > 0 on [0,π) andϕk(π)≥0.
(2) If –ω–11 ≤k< 0, then it follows thatϕk(0)≥0 by Corollary4.3. Thus Proposition4.1 implies the inequalitycos0 = 1≤ (ω–1)kk(ω–2)2–1. On the other hand, we have the inequality cosθ<cos0≤(ω–1)kk(ω–2)2–1 on (0,π]. Thusϕk(θ) > 0 on (0,π] andϕk(0)≥0.
Summing up these cases, we obtain that if 0 <|k| ≤ ω–11 , thenϕk(θ) > 0 on (0,π). Thus, in view of Corollary3.4, we see thatϕk: [0,π]→[0,π] is bijective. Therefore,sinϕk(θ) = 0
if and only ifθ= 0,π.
Remark4.5 If 0 <|k| ≤ ω–11 in Theorem4.4, thenϕk(θ)≥0 on [0,π] andϕk(θ) > 0 (0 <θ<
π),ϕk(0) = 0,ϕk(π) =π.
Finally, we discuss the existence of solutions of EquationBω,k(θ) = 0 on (0,π) for the case ω–11 <|k|< 1.
(1)Properties of the setsZ+(θ)andZ–(θ).
We turn to the existence of solutions in (0,π) of Equation Bω,k(θ) = 0. Clearly, sinϕk(θ) = 0 is reduced to
mπ=ωβ(k,θ) +θ, m∈Z. (17)
Thusm=m(θ,k,ω) on [0,π]. To obtain such anm, we introduce the function βm(θ) = mπ–θ
ω , 0≤θ≤π.
Then (17) is equivalent toβ(k,θ) =βm(θ). Clearly, since|βm(θ)|<π2, we define the set of allm=m(θ,ω)∈Zsatisfying|βm(θ)|<π2.
The following statements are obvious.
Lemma 4.6 Letω–11 <|k|< 1.Then the following statements are equivalent.
(1)For any k,equationBω,k(θ) = 0has a solution in(0,π).
(2)For any k,equationsinϕk(θ) = 0has a solution in(0,π).
(3)For any k,there exists an m∈Zand aθ∈(0,π)satisfyingϕk(θ) =mπ.
(4)For any k there exists an m∈Zand aθ∈(0,π)satisfyingβ(k,θ) =βm(θ).
Fora∈R, the symbol [a] stands for the maximum integer not greater thana. We set ω0= [ω2],O={2n+ 1|n∈Z}andE={2n|n∈Z}. Then we note that ifω∈O, thenω0=
ω
2 –12; ifω∈E, thenω0=ω2. Since|βm(θ)|< π2, it follows that –ω2 +πθ <m<ω2 +πθ. For θ∈(0,π) andω∈Z∞3 , we define
Z+(θ) =
"
m∈Z|1≤m<ω 2 + θ
π
#
, Z–(θ) =
"
m∈Z|–ω 2 +θ
π <m≤0
# ,
andZ(θ) =Z+(θ)∪Z–(θ). Forθ= 0,π, we define
Z(0) =Z–(0) ={0} and Z(π) =Z+(π) ={1}. (18) Next, by easy calculation, we can determine the setsZ+(θ) andZ–(θ) as follows.
Lemma 4.7 Letθ∈(0,π).
(1)Ifω∈O,then
Z+(θ) =
⎧⎨
⎩m∈Z!!!1≤m≤
⎧⎨
⎩
ω0 (0 <θ≤π2) ω0+ 1 (π2 <θ<π)
⎫⎬
⎭,
Z–(θ) =
⎧⎨
⎩m∈Z!!! –ω0(0 <θ<π2) 1 –ω0(π2 ≤θ<π)
#
≤m≤0
⎫⎬
⎭.
(2)Ifω∈E,then
Z+(θ) ={m∈Z|1≤m≤ω0}, Z–(θ) ={m∈Z|1 –ω0≤m≤0}.
Now, we give a relationship between the setZ+(θ) (orZ–(θ)) andβm(θ).
Corollary 4.8 The following statements hold:
(1)m∈Z+(θ)if and only if0 <βm(θ) <π2. (2)m∈Z–(θ)if and only if–π2 <βm(θ) < 0.
The following lemma easily follows from Lemma3.2and Lemma4.7.
Lemma 4.9 The following statements hold.
(1)If0 <k< 1,then there exists an m∈Z+(θ)such that0 <βm(θ) <π2 for allθ∈(0,π).
(2)If–1 <k< 0,then there exists an m∈Z–(θ)such that–π2 <βm(θ) < 0for allθ∈(0,π).
(2)The function gm(θ).
In general, if there exist ak∗, 0 <|k∗|< 1, aθ∗∈[0,π] and anm∗∈Zsuch thatβ(k∗,θ∗) = βm∗(θ∗), thensinϕk∗(θ∗) =sinm∗π= 0. It follows from Corollary4.8thatm∗∈Zcan be replaced bym∗∈Z(θ∗). So, we havem= 0 andm= 1 in (17) for the solutionsθ= 0 and θ=πof the equationsinϕk(θ) = 0.
First, we discuss the existence of solutions of EquationBω,k(θ) = 0, i.e.,sinϕk(θ) = 0.
We define
gm(θ) :=gm,k(θ) =sinβm(θ) –ksin
βm(θ) +θ
, θ∈(0,π), (19)
wherem∈Z(θ). It follows from Corollary4.8and Lemma4.9that if 0 <k< 1, thenm∈ Z+(θ); if –1 <k< 0, thenm∈Z–(θ). We define
gm,k(0) := lim
θ→0+gm,k(θ) = (1 –k)sinmπ ω , gm,k(π) := lim
θ→π–gm,k(θ) = (1 +k)sin(m– 1)π ω . Thengm,k(θ) is well defined on [0,π].
Lemma 4.10 If for any k, ω–11 <|k|< 1,there exist aθ∗∈(0,π)and an m∗∈Z(θ∗)such that gm∗,k(θ∗) = 0,thensinϕk(θ∗) = 0,and vice versa.
Proof Sincegm∗,k(θ∗) = 0, i.e.,sinβm∗(θ∗) =ksin(βm∗(θ∗) +θ∗), (12) yields tanβm∗(θ∗) =
ksinθ∗
1–kcosθ∗. This meansβ(k,θ∗) =βm∗(θ∗). Thussinϕk(θ∗) = 0 by Lemma4.6.
Conversely, let for anyk, ω–11 <|k|< 1, there exist aθ∗∈(0,π) such thatsinϕk(θ∗) = 0.
Then there exists anm∗∈Z(θ) such thatϕk(θ∗) =m∗π. Therefore,β(k,θ∗) =βm∗(θ∗) by Lemma4.6. Thusβm∗(θ∗) satisfies the equationgm∗,k(θ∗) = 0.
The derivative ofgm(θ) becomes gm (θ) = –1
ω
cosβm(θ) +k(ω– 1)cos
βm(θ) +θ
, 0≤θ≤π.